Question

In triangle $ABC$, internal angle bisector angle $A$ makes an angle $\theta$ with side $BC$. The value of sin $\theta$ is equal to:

• A. $|\sin (\frac{B-C}{2})|$
• B. $|\sin (\frac{B}{2}-C)|$
• C. $\cos (\frac{B-C}{2})$
• D. $\cos (\frac{B}{2}-C)$

Answer 1

The correct option is C $\cos (\frac{B-C}{2})$

Consider an isosceles triangle with $AB=AC$.
Now $\angle B=\angle C$.
Now we know that the angle bisector of A will be perpendicular bisector of BC.
Hence,
$\theta ={90}^0$
$\sin \theta =1$
$=\cos \left(0^0\right)$
$=\cos \left(\frac{B-C}{2}\right)$, since $\angle B=\angle C$.
Hence answer is option C.